= Solution
At $(x,0)$ the distances from the two centres are equal:
$$
r=\sqrt{x^2+a^2/4},\qquad x_1=\frac{\sqrt3}{2}a,\qquad x_2=\frac{\sqrt{15}}2a.
$$
By the electrostatic <shell theorem>, a uniformly charged spherical shell gives zero field inside and its point-charge field outside. For $x<x_1$ the point is inside both shells. For $x_1<x<x_2$ only the small, negatively charged shell contributes. For $x>x_2$ both contribute, so
$$
\mathbf E(x)=\begin{cases}
0,&0\le x<x_1,\\
\dfrac{k_eq}{r^3}\left(-x\hat{\mathbf x}+\dfrac a2\hat{\mathbf y}\right),&x_1<x<x_2,\\
\dfrac{k_eqa}{r^3}\hat{\mathbf y},&x>x_2,
\end{cases}\qquad k_e=\frac1{4\pi\epsilon_0}.
$$
The requested magnitude is
$$
\boxed{|\mathbf E(x)|=\begin{cases}
0,&0\le x<x_1,\\
\dfrac{k_eq}{x^2+a^2/4},&x_1<x<x_2,\\
\dfrac{k_eqa}{(x^2+a^2/4)^{3/2}},&x>x_2.
\end{cases}}
$$
It jumps from zero to $k_eq/a^2$ at $x_1$, decreases to $k_eq/(4a^2)$ just before $x_2$, jumps down to $k_eq/(8a^2)$ just after $x_2$, and then falls as $x^{-3}$. The ideal surface charge makes the field discontinuous at each crossing, so the one-sided limits, rather than a unique value on the sheet, specify the graph. The far field is the equatorial field of an <electric dipole> with <electric dipole moment> $\mathbf p=-qa\hat{\mathbf y}$.
\Image[/past-exam-of-the-physics-2-paper-of-the-university-of-oxford/2014-offset-shells.png]
{title=Geometry and electric-field magnitude for the two offset charged shells}
{description=The open endpoints give the one-sided limits at the two surface crossings. The field is zero inside both shells and has a dipolar far-field tail.}
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